Question

AB is a side of a regular hexagon and BC is a side of a regular nonagon. Find ∠BOC, ∠AOB, ∠OBC, ∠OAC and ∠ABC.

Answer 1

Step 1: Find ∠AOB

A regular hexagon has 6 equal sides. The angle subtended by each side at the center of the circumscribed circle is found by dividing the total angle of the circle (360°) by the number of sides.

∠AOB = `360^circ/6`

∠AOB = 60°

Step 2: Find ∠BOC

A regular nonagon has 9 equal sides. The angle subtended by each side at the center of the circumscribed circle is found by dividing the total angle of the circle (360°) by the number of sides.

`∠BOC = 360^circ/9`

∠BOC = 40°

Step 3: Find ∠OBC

In ΔOBC, the sides OB and OC are radii of the circle, so they are equal in length.

OB = OC

Therefore, ΔOBC is an isosceles triangle and the base angles opposite the equal sides are also equal.

∠OBC = ∠OCB

The sum of angles in a triangle is 180°.

∠OBC + ∠OCB + ∠BOC = 180°

2 × ∠OBC + 40° = 180°

2 × ∠OBC = 180° – 140°

2 × ∠OBC = 140°

`∠OBC = 140^circ/2`

∠OBC = 70°

Step 4: Find ∠OAC

In ΔAOC, the sides OA and OC are radii of the circle, so they are equal in length.

OA = OC

Therefore, ΔAOC is an isosceles triangle.

The angle at the center, ∠AOC is the sum of ∠AOB and ∠BOC.

∠AOC = ∠AOB + ∠BOC

= 60° + 40° 

= 100°

The base angles opposite the equal sides are also equal.

∠OAC = ∠OCA

The sum of angles in a triangle is 180°.

∠OAC + ∠OCA + ∠AOC = 180°

2 × ∠OAC + 100° = 180°

2 × ∠OAC = 180° – 100°

2 × ∠OAC = 80°

`∠OAC = 80^circ/2`

∠OAC = 40°

Step 5: Find ∠ABC

In ΔOAB, the sides OA and OB are radii of the circle, so they are equal in length.

OA = OB

Therefore, ΔOAB is an isosceles triangle and the base angles opposite the equal sides are also equal.

∠OAB = ∠OBA

The sum of angles in a triangle is 180°.

∠OAB + ∠OBA + ∠AOB = 180°

2 × ∠OBA + 60° = 180°

2 × ∠OBA = 120°

`∠OBA = 120^circ/2`

∠OBA = 60°

The angle ∠ABC is the sum ∠OBA and ∠OBC.

∠ABC = ∠OBA + ∠OBC

∠ABC = 60° + 70° = 130°

∠ABC = 130°